# Infinitesimal transformations and Poisson brackets

Tags:
1. Feb 11, 2015

Hello, I want to understand how bracket operations in general are related to symmetry and infinitesimal transformations (in hindsight of quantumfieldtheory), so I calculated an example with a particle that is moving on a circle with a generic potential.
(I used simple polar coordinates in two dimensions)

$H(r,p_{r})= \frac{p^{2}_{r}}{2m}+V(r)$
$H(\phi, p_{\phi})=\frac{p^{2}_{\phi}}{2mr^{2}}+V(\phi)$

Now I know that if you take the Poisson bracket with the Hamiltonian you just get the infinitesimal transformation in time right? So
$\{r,H(r,p_{r})\}= \frac{p_{r}}{m}$
$\{\phi,H(\phi,p_{\phi})\}= \frac{p_{\phi}}{mr^{2}}$

But what if I want to do an infinitesimal transformation of the $r$ and $\phi$ coordinates? I know that the generator of translations is just the momentum, and that the generator of rotations is angular momentum. How would I do that with the Poisson bracket in this case? And for example when I do an infinitesimal transformation with the radius $r$, what does that mean? Is it that the radius is infinitesimally transformed, or is it more like a global translation where the whole system is somehow translated? Similarly with the angle $\phi$, is it that the angle is locally changed, or is it that the "whole" system is rotated?

2. Feb 13, 2015

### ch3cooh

Are you referring to this formula
$$\frac{df}{dt} = \{f, H\} + \frac{\partial f}{\partial t}$$
As for infinitesimal transformation, it is just the radius $r$ changes. For example, we may have a infinitesimal canonical transformation
$$Q = q + \alpha G(q,P)$$
where $\alpha$ is infinitesimally small and $G$ is the generator.